# [R] Normal Distribution Quantiles

David Winsemius dwinsemius at comcast.net
Sat Jan 8 16:36:38 CET 2011

On Jan 8, 2011, at 9:25 AM, Rainer Schuermann wrote:

>> Sounds like homework, which is not an encouraged use of the Rhelp
>> list. You can either do it in theory...
>
> It is _from_ a homework but I have the solution already (explicitly
> got that done first!) - this was the pasted Latex code (apologies
> for that, but in plain text it looks unreadable[1], and I thought
> everybody here has his / her favorite Latrex editor open all the
> time anyway...). I'm just looking, for my own advancement and
> programming training, for a way of doing that in R - which, from
>
> I would _not_ misuse the list for getting homework done easily, I
> will not ask "learning statistics" questions here, and I will always
> try to find the solution myself before posting something here, I
> promise!

It would seem to be a straightforward application of the central limit
theorem. Sum of normally distributed IID variables with common
mean ... that sort of thing.  When I simulated it (n=10,000), I got:

50% 80% 90% 95% 99%
100 103 104 105 106

>
> Thanks anyway for the simulation advice,
> Rainer
>
>
>    (4000 - (40*n))   -329
> [1] --------------- = ----
>              1        200
>       (10*(n^-))
>              2

I'm not sure how that expression relates to the problem at hand. But
even at my advanced age, I'm open to education.

--
David.
>
>
>
>
> On Saturday 08 January 2011 14:56:20 you wrote:
>>
>> On Jan 8, 2011, at 6:56 AM, Rainer Schuermann wrote:
>>
>>> This is probably embarrassingly basic, but I have spent quite a few
>>> hours in Google and RSeek without getting a clue - probably I'm
>>>
>>> There is this guy who has decided to walk through Australia, a total
>>> distance of 4000 km. His daily portion (mean) is 40km with an sd of
>>> 10 km. I want to calculate the number of days it takes to arrive
>>> with 80, 90, 95, 99% probability.
>>> I know how to do this manually, eg. for 95%
>>> $\Phi \left( \frac{4000-40n}{10 \sqrt{n}} \right) \leq 0.05$
>>> find the z score...
>>>
>>> but how would I do this in R? Not qnorm(), but what is it?
>>
>> Sounds like homework, which is not an encouraged use of the Rhelp
>> list. You can either do it in theory or you can simulate it. Here's a
>> small step toward a simulation approach.
>>
>>> cumsum(rnorm(100, mean=40, sd=10))
>>   [1]   41.90617   71.09148  120.55569  159.56063  229.73167
>> 255.35290  300.74655
>> snipped
>>  [92] 3627.25753 3683.24696 3714.11421 3729.41203 3764.54192
>> 3809.15159 3881.71016
>>  [99] 3917.16512 3932.00861
>>> cumsum(rnorm(100, mean=40, sd=10))
>>   [1]   38.59288   53.82815  111.30052  156.58190  188.15454
>> 207.90584  240.64078
>> snipped
>>  [92] 3776.25476 3821.90626 3876.64512 3921.16797 3958.83472
>> 3992.33155 4045.96649
>>  [99] 4091.66277 4134.45867
>>
>> The first realization did not make it in the expected 100 days so
>> further efforts should extend the simulation runs to maybe 120 days.
>> The second realization had him making it on the 98th day. There is an
>> R replicate() function available once you get a function running that
>> will return a specific value for an instance. This one might work:
>>> min(which(cumsum(rnorm(120, mean=40, sd=10)) >= 4000) )
>> [1] 97
>>
>> If you wanted a forum that does not explicitly discourage homework
>> and
>> would be a better place to ask theory and probability questions,
>> there
>> is CrossValidated:
>> http://stats.stackexchange.com/faq
>>
>>>